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\textsc{中国科学院大学   计算机与控制学院} \\ [25pt] % Your university, school and/or department name(s)
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\huge 随机过程第九次作业 \\ % The assignment title
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\author{黎吉国&201618013229046} % Your name
%\date{\normalsize Nov 8,2016}

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\section{过滤泊松过程}
设$\{ X(t),0\le t < \infty \}$，并有
\[ X(t)=\Sigma_{j=0}^{N_t} h(t-U_j) \]
式中，在时刻$U_j$发生的事件，在时刻$t$的输出为$h(t-U_j)$，式中在间隔$(0,t)$内发生的事件
由泊松计数随机变量$N$描述，随机变量$U_j$是在间隔$(0,t)$中发生事件的无序到达时刻，这种过程称为过滤泊松随机过程。
求特征函数$\phi_{X_t}(v)$.\\
\textbf{解：}\\
$X(t)$中有两个变量具有随机性:$U_j,N_t$,利用条件期望
\[ E(Y)=E_X(E_Y(Y|X)) \]
根据特征函数的定义可得
\[
\begin{split}
  \phi_{X_t}(v) &= E\{ e^jvX(t) \}\\
  &= E_{N_t}\{ E_{U_j}(e^{jvX(t)}|N_t) \}\\
  &= \Sigma_{k=0}^{\infty} E(e^{jvX(t)|N_t = k})P(N_t=k)\\
\end{split}
\]
这里
\[ E(e^{jvX(t)}|N_t=k)=E(\exp( jv\Sigma_{i=1}^{k} h(t-U_i) )) \]\\
由于$U_i$是统计独立且具有同分布的随机变量，故有
\[ E( e^{jvX(t)}|N_t=k )=\prod_{i=1}^{k}E\{ \exp(jvh(t-U_i)) \}=\{ E(\exp(jvh(t-U_i))) \}^k \]
因为$U_i$是在$(0, t)$内是均匀分布的，所以
\[
\begin{split}
  E\{ \exp(jvh(t-U_i)) \} &= \frac{1}{t}\int_{0}^{1}\exp(jvh(t-u_i))d u_i\\
  &= \frac{1}{t} \int_0^t \exp(jvh(u))du
\end{split}
\]
将结果代入$\phi_{X_t}(v)$，可得
\[
\begin{split}
  \phi_{X_t}(v) &= \Sigma_{k=0}^{\infty}\{ \frac{1}{t}\int_{0}^{t}\exp (jvh(u))du \}^k \frac{(\lambda t)^k}{k!}e^{-\lambda t}\\
  &= e^{-\lambda t}\exp \{ \lambda \int_{0}^{t}\exp(jvh(u))du \}\\
  &= \exp\{ \lambda \int_{0}^{t}( \exp\{ jvh(u) \}-1 )du \}
\end{split}
\]

\section{泊松过程}
对于上题所定义的泊松过程
\[  X(t)=\Sigma_{j=1}^{N_t}h(t-U_j,Y_j) \]
考虑特殊情况
\[
h(\tau,Y_j)=
\begin{cases}
  1 &\mbox{ $0\le\tau <Y_j$}\\
  0 &\mbox{其它}
\end{cases}
\]
其中$Y_j$是统计独立，分布相同的非负随机变量。设
\[ \phi_{X_t}(v) = \lambda\int_0^t \{ 1-F_Y(s) \}ds \]
\begin{enumerate}
  \item 证明$X(t)$是泊松随机变量，其均值是
  \[ E(X(t))=\lambda \int_0^t (1-F_Y(s))ds  \]
  \item 证明
  \[ \lim_{t\to \infty}E(X(t))=\lambda E(Y) \]
\end{enumerate}
\textbf{证明：}\\
\begin{enumerate}
  \item 由于
  \[ h(\tau,Y)=
  \begin{cases}
    1 &\mbox{$0\le \tau <Y$}\\
    0 &\mbox{其它}
  \end{cases}
   \]
   所以
   \[
   E\{ e^{jv h(\tau,Y)}-1 \}= (e^{jv}-1)P\{ Y\ge\tau \}=(e^jv-1)(1-F_Y(\tau))
   \]
   代入$ \phi_{X_t}(v) $,则有
   \[ \phi_{X_i}(v)=\exp \{ \lambda \int_0^{t}(1-F_Y(\tau))(e^{jv}-1)d\tau \} \]
   故
   \[ E\{ X(t) \}=-j \frac{d \phi_{X_t}(v)}{dv}|_{v=0}=\lambda \int_0^t (1-F_Y(\tau))d\tau \]
   \item 因为
   \[
   \begin{split}
   \lambda \int_0^\infty (1-F_Y(\tau)) d\tau  &= \lambda(1-F_Y(\tau))|_0^\infty + \lambda \int_0^\infty \tau dF_Y(\tau)\\
   &= \lambda\int_0^\infty \tau p_Y(\tau)d\tau\\
   &= \lambda E(Y)
 \end{split}
   \]
   所以
   \[ \lim_{t\to \infty} E(X(t))=\lambda\int_0^\infty (1-F_Y(\tau))d\tau=\lambda E(Y) \]
\end{enumerate}
\section{过滤泊松过程应用}
到达一台交换机（具有无数通道）的呼叫电话复合参数为$\lambda$的泊松过程，每次对话服从平均值为$\frac{1}{\mu}$的指数分布.
令$X(t)$是$t$时刻对话的数目。
\begin{enumerate}
  \item 求$E(X(t))$和$\lim_{t\to \infty}E(X(t))$;
  \item 求$Var(X(t))$和$\lim_{t\to \infty}Var(X(t))$;
  \item 在过程中没有呼叫的概率是多少？
\end{enumerate}
\textbf{解：}\\
\begin{enumerate}
  \item 根据题意，有
  \[
  p_Y(\tau)=
  \begin{cases}
  \mu e^{-\mu x} &\mbox{$\tau \ge 0$}\\
  0 &\mbox{$\tau<0$}
  \end{cases}
  \]
  利用上题的结论，可得
  \[
  \begin{split}
    E(X(t))&=\lambda\int_0^t (1-F_Y(\tau))d\tau\\
    &= \lambda\int_0^t e^{-\mu\tau}d\tau\\
    &= \frac{\lambda}{\mu}(1-e^{-\mu\tau})\\
    \lim_{t\to \infty}E(X(t))&=\frac{\lambda}{\mu}
  \end{split}
  \]
  \item 因为泊松随机变量的均值和方差是相同的，所以有
  \[ var(X(t))=\frac{\lambda}{\mu}(1-e^{-\mu t}) \]
  \[ \lim_{t\to \infty} var(X(t))=\frac{\lambda}{\mu} \]
  \item 没有呼叫意味着$X(t)=0$,故没有呼叫的概率为
  \[P(X(t)=0)=\exp\{ -E(x(t)) \}=\exp \{ -\frac{\lambda}{\mu}(1-e^{-\mu t}) \}\]
\end{enumerate}

\section{总结}
以上三题从泊松随机过程的定义出发，分析了其均值方差等性质，还给出了具体的应用有助于我们掌握过滤随机过程的本质，
能够帮助我们更好地分析实际的问题。


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